jee(main) – 2015 test paper with solution (held on...
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1
1. Distance of the centre of mass of a solid uniform
cone from its vertex is z0. If the radius of its base
is R and its height is h then z0 is equal to :-
(1) 5h
8(2)
23h
8R
(3) 2h
4R(4)
3h
4
Ans. (4)
Sol.
h4
for solid cone c.m. is h
4 from base
so z0 = h – h
4=
3h
4
2. A red LED emits light at 0.1 watt uniformly
around it. The amplitude of the electric field of
the light at a distance of 1 m from the diode is:-
(1) 5.48 V/m (2) 7.75 V/m
(3) 1.73 V/m (4) 2.45 V/m
Ans. (4)
Sol. Iav = 1
2 0E2C =
2
P
4 r
E = 2 0
2P
4 r c
On putting value we get
= 2.45 v/m.
PART A – PHYSICS
JEE(MAIN) – 2015 TEST PAPER WITH SOLUTION(HELD ON SATURDAY 04th APRIL, 2015)
3. A pendulum made of a uniform wire of cross
sectional area A has time period T. When an
additional mass M is added to its bob, the time
period changes to TM. If the Young's modulus
of the material of the wire is Y then 1
Y is equal
to :- (g = gravitational acceleration)
(1) 2
MT A1
T Mg
(2)
2
M
T A1
T Mg
(3)2
MT A1
T Mg
(4) 2
MT Mg1
T A
Ans. (3)
Sol. T = 2L
g...(1)
TM = 2L L
g...(2)
Y =
FA
LL
L = FL
AY...(3)
Putting (3) in (2)
solving the equation we get the value of 1
Y as
2
MT A–1
T mg
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2
4. For a simple pendulum, a graph is plotted
between its kinetic energy (KE) and potential
energy (PE) against its displacement d. Which
one of the following represents these
correctly ? (graphs are schematic and not
drawn to scale)
(1)
E
KE
d
PE
(2)
PE
KE
E
(3) PE
KE
E
d
(4)
PE
KE
E
d
Ans. (4)
Sol. KE = 1
2m2[A2 – d2], downward parabola
PE = 1
2m2d2, upward parabola.
KE
PE
5. A train is moving on a straight track with speed
20 ms–1. It is blowing its whistle at the
frequency of 1000 Hz. The percentage change
in the frequency heard by a person standing
near the track as the train passes him is
(speed of sound = 320 ms–1) close to :-
(1) 18% (2) 24% (3) 6% (4) 12%
Ans. (4)
Sol. f' = S
vf
v v
f" = S
vf
v v
f "
f ' =
S
S
v v
v v
f " f '100
f '
=
S S
S
v v v v10
v v
= S
S
2v
v v
=
S
2
v1
v
× 100
~ 12%
6. When 5V potential difference is applied across
a wire of length 0.1 m, the drift speed of
electrons is 2.5 × 10–4 ms–1. If the electron
density in the wire is 8 × 1028 m–3, the resistivity
of the material is close to :-
(1) 1.6 × 10–6 m (2) 1.6 × 10–5 m
(3) 1.6 × 10–8 m (4) 1.6 × 10–7 m
Ans. (2)
Sol. V = iR =
d(neAV )
A
V= neVd
d
V
neV
On putting values are got the answer
= 1.6 × 10–5 m
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3
7.
L
II
Two long current carrying thin wires, both with
current I, are held by the insulating threads of
length L and are in equilibrium as shown in the
figure, with threads making an angle '' with
the vertical. If wires have mass per unit length
then the value of I is :-
(g = gravitational acceleration)
(1) 0
gL2 tan
µ
(2)
0
gLtan
µ
(3) 0
gLsin
µ cos
(4)
0
gL2sin
µ cos
Ans. (4)
B = 0µ I
2 r =
0µ I
2 2 sin
tan = Bf
g where fB is force per unit length(Bi)
g tan = 0µ I
I2 2 sin
on solving
I = 2 sin0
g
µ cos
8.
6V
9V
2P
1
Q 33
In the circuit shown, the current in the 1resistor is :-(1) 0.13 A, from Q to P(2) 0.13 A, from P to Q(3) 1.3 A, from P to Q(4) 0A
Ans. (1)
Sol.
0V 9V3R
6VR 5R
P
Q
E =
6 0 9
3 1 51 1 1
3 1 5
=
92
55 15 3
15
= 3
23
i = E
R =
3
23 = 0.13
from +ve to –veQ to P
9. Assuming human pupil to have a radius of
0.25cm and a comfortable viewing distance of
25cm, the minimum separation between two
objects that human eye can resolve at 500 nm
wavelength is :-
(1) 100 µm (2) 300 µm
(3) 1 µm (4) 30 µm
Ans. (4)
Sol.0
1
d 1.22
d d
92
0 2
1.22 500 10d 25 10
0.5 10
d0 = 1.22 × 25 × 10–6
d0 30 µm
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4
10. An inductor (L = 0.03 H) and a resistor
(R = 0.15 k) are connected in series to a
battery of 15V EMF in a circuit shown below.
The key K1 has been kept closed for a long
time. Then at t = 0, K1 is opened and key K2
is closed simultaneously. At t = 1ms, the current
in the circuit will be (e5 150) :-
K2
K1
15V
0.03H 0.15k
(1) 6.7 mA (2) 0.67 mA
(3) 100 mA (4) 67 mA
Ans. (2)
Decay of current
tR
L0I I e
=
310 150
0.0315
e150
= 51
e10
= 1
1500
= 6.66 × 10–4
= 0.666 × 10–3
= 0.67 mA
11. An LCR circuit is equivalent to a damped
pendulum. In an LCR circuit the capacitor is
charged to Q0 and then connected to the L and
R as shown below :-
C
LR
If a student plots graphs of the square of
maximum charge 2Max(Q ) on the capacitor with
time (t) for two different values L1 and L2
(L1 > L2) of L then which of the following
represents this graph correctly ? (plots are
schematic and not drawn to scale)
(1)
L1
L2
Q2
Max
t
(2)
Q2
Max
t
Q (For both L and L )0 1 2
(3)
L1
L2
Q2
Max
t
(4) L1
L2Q2
Max
t
Ans. (3)
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5
(3)
3µF1µFC
Charge
Q2
(4)
3µF1µFC
Charge
Q2
Ans. (4)
Sol.
Ceq = 3C
3 CQ = Ceq.E
Q2 = 2CE
C 3 = 3CE
C 3
= 2E3
1C 3
Q2/when C = 1µF = 2E 1
4
= E
2
Q2/when C = 3µF = 2E 1
2
= E
dQ
dC =
2E 3
1C 3
= 2
6E0
C 3
2
2
d
dC
=
3
6E 2
C 3
=
2
12E0
C 3
Sol. As damping is happening its amplitude wouldvary as
Q0
The oscillations decay exponentially and willbe proportional to e–t where dependsinversely on L.So as inductance increases decay becomesslower for
L1 L2
12. In the given circuit, charge Q2 on the 2µFcapacitor changes as C is varied from 1µF to3µF. Q2 as a function of 'C' is given properlyby : (figures are drawn schematically and arenot to scale) :-
1µF
2µF
C
E
(1)
3µF1µFC
Charge
Q2
(2)
3µF1µFC
Charge
Q2
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6
13. From a solid sphere of mass M and radius R a
cube of maximum possible volume is cut.
Moment of inertia of cube about an axis
passing through its centre and perpendicular to
one of its faces is:-
(1)
24MR
9 3(2)
24MR
3 3
(3)
2MR
32 2(4)
2MR
16 2
Ans. (1)
Sol.
R
R
a
Let mass and side of cube be M' and a 3 a = 2R
M' = 3
3
Ma
42R
3
Moment of Inertia of cube = 2M 'a
6
=
2 23
3
M a 4MRa
4 6 4 32R3
14. The period of oscillation of a simple pendulum
is T = L
2g
. Measured value of L is 20.0 cm
known to 1 mm accuracy and time for 100oscillations of the pendulum is found to be 90susing a wrist watch of 1s resolution. Theaccuracy in the determination of g is :
(1) 1% (2) 5%
(3) 2% (4) 3%
Ans. (4)
Sol. T = 2
g
g = 2
2
4
T
g T2
g T
–3
–2
g 1 10 12
90g 20 10 100100
g
g × 100 = 2.722% 3%
15. On a hot summer night, the refractive index of
air is smallest near the ground and increases
with height from the ground. When a light beam
is directed horizontally, the Huygens' principle
leads us to conclude that as it travels, the light
beam :
(1) bends downwards
(2) bends upwards
(3) becomes narrower
(4) goes horizontally without any deflection
Ans. (2)
Sol. Normalµµ
2
1
µ increases
2 > 1
light bends towards normal
light beam bends upwards (as µwith height)
16. A single of 5 kHz frequency is amplitude
modulated on a carrier wave of frequency
2 MHz. The frequencies of the resultant signal
is/are -
(1) 2005 kHz, 2000 kHz and 1995 kHz
(2) 2000 kHz and 1995 kHz
(3) 2 MHz only
(4) 2005 kHz and 1995 kHz
Ans. (1)
Sol. Frequency present after modulationf
c, f
c ± f
s
2000 KHz, 2005 KHz and 1995 KHz17. A solid body of constant heat capacity 1 J°/C
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is being heated by keeping it in contact withreservoirs in two ways -(i) Sequentially keeping in contact with 2
reservoirs such that each reservoir suppliessame amount of heat.
(ii) Sequentially keeping in contact with 8reservoirs such that each reservoir suppliessame amount of heat.
In both the cases body is brought from initialtemperature 100°C to final temperature 200°C.Entropy change of the body in the two casesrespectively is -
(1) ln2, 2ln2 (2) 2ln2, 8ln2
(3) ln2, 4ln2 (4) ln2, ln2
Ans. (4)
Sol.2
1
T
T
dTS nc
T
200
1 100
dTS 1 n2
T
200
2 100
dTS 1 n2
T
18. Consider a spherical shell of radius R attemperature T. The black body radiation insideit can be considered as an ideal gas of photonswith internal energy per unit volume
u = 4U
TV and pressure p =
1 U
3 V
. If the shell
now undergoes an adiabatic expansion therelation between T and R is -
(1) T 1
R(2) T 3
1
R
(3) T e–R (4) T e–3R
Ans. (1)
Sol. P = 1 U
3 V
P T4
using PV = nRT
31T
V
T 1
R
19. Two stones are thrown up simultaneously from
the edge of a cliff 240 m high with initial speed
of 10 m/s and 40 m/s respectively. Which of
the following graph best represents the time
variation of relative position of the second
stone with respect to the first ?
(Assume stones do not rebound after hitting the
ground and neglect air resistance, take g =
10 m/s2)
(The figure are schematic and not drawn to
scale)
(1)
240(y –y )m2 1
8 12t(s)
(2)
240(y –y )m2 1
8 12t(s)
(3)
240(y –y )m2 1
8 12t(s)
t
(4)
240(y –y )m2 1
12t(s)
Ans. (1)
Sol. For particle 1
A
B
240m
10m/s
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–240 = + 10t – 2
gt2
5t2 – 10t – 240 = 0t
1 = 8 sec
For particle 2
A
B
40m/s
–240 = 40t – 2
gt2
5t2 – 40t – 240 = 0t
2 = 12 sec
for 0 < t < 8 sec arel
= 0straight line x-t graph
for 8 < t < 12 sec arel
= – gdownward parabola
for t > 12 sec Both particles comes to rest20. A uniformally charged solid sphere of radius
R has potential V0 (measured with respect to )
on its surface. For this sphere the equipotential
surfaces with potentials 0 0 03V 5V 3V
, ,2 4 4
and 0V
4
have radius R1 , R2 , R3 and R4 respectively. Then
(1) R1 = 0 and R2 < (R4 – R3)
(2) 2R < R4
(3) R1 = 0 and R2 > (R4 – R3)
(4) R1 0 and (R2 – R1) > (R4 – R3)
Ans. (1 or 2)
Sol. for r < R2
2
kQ rV 3
2R R
& for r > RkQ
Vr
at r = R 0
kQV
R
at r = 0 0 0
3 kQ 3V V
2 R 2
R1 = 0
for 0
5V V
4
2
2
kQ r 5 kQ3 .
2R R 4 R
2
RR
2
for 0
3V V
4
3
kQ 3 kQ
R 4 R
3
4RR
3
for 0VV
4
4 0
kQ kQ
R 4R
R4 = 4R0 .
21. Monochromatic light is incident on a glass
prism of angle A. If the refractive index of the
material of the prism is µ, a ray, incident at an
angle , on the face AB would get transmitted
through the face AC of the prism provided :
C
A
B
(1) 1 1 1
cos µ sin A sinµ
(2) 1 1 1cos µ sin A sin
µ
(3) 1 1 1sin µ sin A sin
µ
(4) 1 1 1sin µ sin A sin
µ
Ans. (3)
Sol.
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For all light to come out from face AC angle
of emergence e 90
Apply Snell's Law at face AC
µ sin r2 = 1 sin e
r2 = sin–11
µ
(if e = 90)
r1 = A – sin–11
µ
( r1 + r2 = A)
Apply Snell's law at face AB
1 sin = µsin(r1)
= -1 -1 1
sin µ sin(A-sin µ
for all light transmitted through AC, e < 90°
-1 -1 1> sin µ sin(A-sin
µ
22. A rectangular loop of sides 10 cm and 5 cm
carrying a current I of 12 A is place in different
orientations as shown in the figures below :
(a)
z
y
x
I
I
IB
I
(b)
z
y
x I
B
I
II
(c)
z
y
x
B
II
I
I
(d)
z
y
x I
B
I
II
If there is a uniform magnetic field of 0.3 T in
the positive z direction, in which orientations
the loop would be in (i) stable equilibrium and
(ii) unstable equilibrium ?
(1) (b) and (d), respectively
(2) (b) and (c), respectively
(3) (a) and (b), respectively
(4) (a) and (c), respectively
Ans. (1)
Sol. For stable equilibrium B M
For unstable equilibrium B = magnetic fieldM = magnetic moment
23. Two coaxial solenoids of different radii carry
current I in the same direction. Let 1F
be the
magnetic force on the inner solenoid due to the
outer one and 2F
be the magnetic force on the
outer solenoid due to the inner one. Then :
(1) 1F
is radially inwards and 2F 0
(2) 1F
is radially outwards and 2F 0
(3) 1 2F F 0
(4) 1F
is radially inwards and 2F
is radially
outwards.
Ans. (3)
Sol.
S1
S2
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Field due to solenoid 2 outside the solenoid iszero
1f 0
Due to action reaction 2f 0
or solenoid 2 behave as magnetic dipole
2
Md Bf
dx
field of 1 is uniform dB
0dx
f2 = 0
24. A particle of mass m moving in the x direction
with speed 2is hit by another particle of mass
2m moving in the y direction with speed . If the
collisions perfectly inelastic , the percentage loss
in the energy during the collision is close to :
(1) 56 % (2) 62%
(3) 44% (4) 50%
Ans. (1)Sol. Before collison
m 2v
2m
v
y
x
Kinetic energy = 2 21 1m(2v) 2m(v)
2 2 = 3 mv2
After collisonApplying momentum conservation for inelasticcollision
fˆ ˆ2mvj m2vi 3m v
f
8vv
9
2
2f f
1 4mvK 3m v
2 3
% K = 2
i f2
i
K – K 5mv / 3 556%
K 3mv 9
25. Consider an ideal gas confined in an isolated
closed chamber. As the gas undergoes an
adiabatic expansion, the average time of
col l ision between molecules increases as Vq,
where V is the volume of the gas. The value
of q is :-
p
v
C
C
(1)1
2
(2)
1
2
(3) 3 5
6
(4)
3 5
6
Ans. (1)
Sol. avg
Volume
v
1
21V T
For adiabatic processT V1–
1
1 –1/ 2 2V T V
comparing
1q
2
26. From a solid sphere of mass M and radius R,
a spherical portion of radius R
2 is removed, as
shown in the figure. Taking gravitational
potential V = 0 at r = , the potential at the
centre of the cavity thus formed is :
(G = gravitational constant)
(1) 2GM
3R
(2)
2GM
R
(3) GM
2R
(4)
GM
R
Ans. (4)
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11
Sol. By principle of superosition
22
3
GM 3G MRV – 3R –
R22R 4 82
–11 GM GM GM
–8R 8R R
27. BAF
Given in the figure are two blocks A and B of
weight 20 N and 100 N, respectively. These are
being pressed against a wall by a force F as
shown. If the coefficient of friction between the
blocks is 0.1 and between block B and the wall
is 0.15, the frictional force applied by the wall
on block B is :-
(1) 120 N (2) 150 N
(3) 100 N (4) 80 N
Ans. (1)Sol.
FA
20
B 100
100f1
f1 f2
for equllibrrium of Af1 = 20
for equllibrrium of Bf2 = f1 + 100f2 = 120 N
28. A long cylindrical shell carries positive surfacecharge in the upper half and negative surfacecharge – in the lower half. The electric fieldlines around the cylinder will look like figuregiven in : (figures are schematic and not drawnto scale)
(1)+ ++
+++ +––––––
(2)
+++––– –––
++–
(3)++
+––– –––
++–
(4)++
+––– –––
++–
Ans. (3)
It behaves as a dipole.
++––– –––
+–
+
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29. As an electron makes a transition from an
excited state to the ground state of a
hydrogen - like atom/ion :
(1) kinetic energy decreases, potential energy
increases but total energy remains same
(2) kinetic energy and total energy decrease but
potential energy increases
(3) its kinetic energy increases but potential
energy and total energy decreases
(4) kinetic energy, potential energy and total
energy decrease
Ans. (3)
Sol. K = 2
2
z13.6
n as n decreases k increases
U = 2
2
z27.2
n
as n decrease
U & T decrease
T = 2
2
z13.6
n
30. Match List-I (Fundament Experiment) with
List-II (its conclusion) and select the correct
option from the choices given below the list :
List-I List-II
(A) Franck-HertzExperiment.
(i) Particlenature of light
(B) Photo-electricexperiment
(iii) Discreteenergy levelsof atom
(C) Davison-GermerExperiment
(iiii) Wave natureof electroc
(iv) Structure ofatom
(1) A-ii, B-i, C-iii (2) A-iv, B-iii, C-ii
(3) A-i, B-iv, C-iii (4) A-ii, B-iv, C-iii
Ans. (1)
Sol. Self Explanatory/Theory
(A)Franck-Hertz experiment explains disrete
energy levels of atom
(B) Photo-electric experiment explain particle
nature of light
(C) Davison Germer experiment explain wave
nature of electron.
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